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Hoffmann Bromamide degradation converts a primary amide into a primary amine with one less carbon atom using:
$$Br_2/NaOH$$
Option A involves direct conversion of an amide into an amine using Hoffmann degradation.
Option B first forms benzamide by hydrolysis of benzonitrile and then converts it into aniline using Hoffmann degradation.
Option D first forms benzamide from benzoyl chloride and then converts it into aniline using Hoffmann degradation.
In Option C, the first step is the haloform reaction of a methyl ketone with:
$$Br_2/NaOH$$
The resulting acid is converted into an amide and finally reduced using:
$$LiAlH_4$$
Reduction with $$LiAlH_4$$ converts the amide into an amine without loss of carbon and does not involve Hoffmann Bromamide degradation.
Therefore, Option C does not involve Hoffmann Bromamide degradation.
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